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Question

# One mole of an ideal gas (monatomic) is taken from A to C along path ABC. The temperature of gas is T0 at A, then for process ABC

A
Work done by gas is RT0.
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B
Change in internal energy is 112RT0.
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C
Heat absorb by gas is 112RT0.
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D
Heat absorb by gas is 132RT0.
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Solution

## The correct options are A Work done by gas is RT0. C Heat absorb by gas is 112RT0.Area under PV graph= work done=P0V0=RT0 P0V0T0=2P0.2V0TCâŸ¹TC=4T0 Since, Î”U=nCVÎ”T Î”U=1Ã—32R(4T0âˆ’T0)Î”U=92RT0Q=Î”U+WBy putting valuesQ=112RT0

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